B Fewer seconds or shorter seconds?

[Kairos]

In the twin experiment, the travel time is shorter for the traveling brother than for the static brother. Since the unit of time in physics is the second, is the travel duration shorter because: (1) it contains fewer seconds or (2) the number of seconds is the same for both brothers but the traveler's seconds are shorter?

 

[Ibix]

If I draw a triangle and have two brothers start at one corner and walk the perimeter, meeting at a third corner, one brother walks two sides and the other only one side. The distance walked along two sides is greater than along the third. Is this because: (1) it contains more meters or (2) the number of meters is the same for both brothers but the two-side brother's meters are shorter?

The twin paradox is just this same phenomenon but in a Minkowski space not a Euclidean one. In Minkowski geometry the straight line between two (timelike separated) points is the longest distance (or interval, to give it its technical name).

 

[mitochan]

(1); The traveling twin has experienced fewer seconds so has had fewer birthdays in long journey than his Earth brother.

 

[Kairos]

(1); The traveling twin has experienced fewer seconds

Ok the second is unchanged but on arrival, there will be less counted on the traveler's quartz watch. Thank you I will meditate on this result!

 

[Ibix]

Ok the second is unchanged but on arrival, there will be less counted on the traveler's quartz watch. Thank you I will meditate on this result!

Rather than meditating I recommend drawing Minkowski diagrams, marking the seconds (or years or whatever) along each path, in the stay-at-home frame and the traveller's inbound and outbound frames. In fact, if you look up posts by former poster @ghwellsjr you will probably find exactly that fairly quickly.

 

[mitochan]

Ok the second is unchanged but on arrival, there will be less counted on the traveler's quartz watch. Thank you I will meditate on this result!

You're welcome.

From Wikipedia-Second

The second is defined as being equal to the time duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom.[1][2]

The twin brothers share this definition of time. The traveling brother has his Ce133 clock which shows his second. The Earth brother has his Ce133 clock which shows his second.

 

[Kairos]

Rather than meditating I recommend drawing Minkowski diagrams

These diagrams are graduated using the letter "t" but it was not clear to me whether dt > dt' corresponds to a dilation of the second or an increase in the number of seconds.

 

[Ibix]

These diagrams are graduated using the letter "t" but it was not clear to me whether dt > dt' corresponds to a dilation of the second or an increase in the number of seconds.

I'm struggling to understand what you mean. The twin paradox is exactly what I said it is in #2 - two paths which have different "lengths" and form a triangle. In Euclidean space I would not describe the difference in path lengths as due to either "a dilation of the meter" or "an increase in the number of meters". Nor would I use your terminology to describe the exact same thing in Minkowski spacetime. It's just two paths of different lengths.

 

[Ibix]

Or perhaps you are talking about time dilation, rather than the differential aging phenomenon in the twin paradox?

 

[vanhees71]

These diagrams are graduated using the letter "t" but it was not clear to me whether dt > dt' corresponds to a dilation of the second or an increase in the number of seconds.

The second is defined in a specific frame of reference, namely the rest frame of the Cs-133 atoms and the frequency of the hyperfine transition as measured in this reference frame.

 

[Dale]

(2) the number of seconds is the same for both brothers but the traveler's seconds are shorter?

Shouldn't that be “longer” instead of “shorter”?

 

[Nugatory]

Shouldn't that be “longer” instead of “shorter”?

Wonderful stuff, this natural language...
It depends on whether a "shorter" second is one during which we age less, or one that it takes more of to fill the gap between the separation and arrival times. Like one of those reversing perspective line drawings of a cube, I can read it either way.

 

[Kairos]

Shouldn't that be “longer” instead of “shorter”?

sorry for the delay. No because n longer time units would have given a longer total duration whereas the traveller is younger on arrival.. but there is no need to argue about this point given the universality of the second

 

[Kairos]

Or perhaps you are talking about time dilation, rather than the differential aging phenomenon in the twin paradox?

no, the time dilation of the SR is reciprocal and not sufficient to explain the age difference in the twins' experiment

 

[Kairos]

The second is defined in a specific frame of reference, namely the rest frame of the Cs-133 atoms and the frequency of the hyperfine transition as measured in this reference frame.

OK and I suppose that a unit of time defined using the cesium 133 or any other atomic transition, is also the same as measured in the reference frame of other contexts (acceleration, gravity) ?

 

[PeroK]

no, the time dilation of the SR is reciprocal and not sufficient to explain the age difference in the twins' experiment

Time dilation is rarely sufficient to explain anything. It's quite common when learning SR to learn time dilation first and then, armed with that alone, try to explain certain phenomena. Time dilation is not enough: in SR you have time dilation, length contraction and the relativity of simultaneity. These three are encapsulated by the Lorentz Transformation. - and generally you need all three to explain anything.

The relativity of simultaneity is so important that, personally, I would teach that first so that hopefully nobody tries to do anything using only time dilation.

 

[Dale]

sorry for the delay. No because n longer time units would have given a longer total duration whereas the traveller is younger on arrival.. but there is no need to argue about this point given the universality of the second

I am about 6 feet tall or about 2 m tall. So according to you a m is shorter than a foot?

 

[Kairos]

according to me, 6 m is taller than 6 feet because m is a longer unit..

 

[Dale]

according to me, 6 m is taller than 6 feet because m is a longer unit..

So you agree that a m is longer than a foot.

Then if two people measure my height, one in m and one in ft, then the person measuring my height in m will get a smaller number than the person measuring my height in ft. So the smaller number corresponds to the larger unit.

the number of seconds is the same for both brothers but the traveler's seconds are shorter?

The traveller records a smaller number so it must be a larger unit, yes? If not, I don’t know at all what you are talking about in your OP at all.

 

[vanhees71]

OK and I suppose that a unit of time defined using the cesium 133 or any other atomic transition, is also the same as measured in the reference frame of other contexts (acceleration, gravity) ?

In other reference frames, where the emitting atom is moving, the spectral lines are Doppler shifted. If gravity is considered, you have to use General Relativity to calculate the corresponding shift of the frequency of the emitted em. waves, and this is closely related to relativistic time dilatation. When calculated with GR, it contains both the Doppler and the gravitational shifts of the frequency.

The twin paradox is something slightly different. Here you compare the proper time between the departure of the twins from each other and compare their proper times when meeting again. These proper times of each twin, i.e., there aging between departure and meeting again, are what clocks traveling with each of the twins measure, and it's a scalar quantity, i.e., independent of any space-time coordinates used to describe the motion of the twins and calculating their proper times. Such and only such quantities are of true and unambigous physical significance.

 

[Kairos]

So you agree that a m is longer than a foot.

Then if two people measure my height, one in m and one in ft, then the person measuring my height in m will get a smaller number than the person measuring my height in ft. So the smaller number corresponds to the larger unit.

There is a misunderstanding, please read my initial question. The duration of the trip is shorter for the traveler than for the sedentary. I was just considering two possibilities for this: (1) they did not count the same number of seconds during the travel interval or (2) they would have counted the same number of seconds without realizing that a second would not have the same value. But the question, certainly stupid, is definitely answered: the first hypothesis is right, the second hypothesis wrong. Thanks

 

[Kairos]

In other reference frames, where the emitting atom is moving, the spectral lines are Doppler shifted. If gravity is considered, you have to use General Relativity to calculate the corresponding shift of the frequency of the emitted em. waves, and this is closely related to relativistic time dilatation. When calculated with GR, it contains both the Doppler and the gravitational shifts of the frequency.

OK thanks. My question was in the same reference frame. Whether cesium 133 is far from any mass or in a very intense gravitational field, the second defined using this cesium in its reference frame will always be exactly the same, is it correct?

 

[PeroK]

OK thanks. My question was in the same reference frame. Whether cesium 133 is far from any mass or in a very intense gravitational field, the second defined using this cesium in its reference frame will always be exactly the same, is it correct?

The simple answer is yes.

A fuller answer is that the theory of GR predicts that proper time (as measured by a perfect clock) is a measure of spacetime distance. You can take cesium 133 as a definitive measure of time locally.

Take two laboratories starting out together then taking different paths through spacetime. In each laboratory local experiments have the expected outcomes: the half life of radioactive elements; the behaviour of a cesium clock and anything else. If the two laboratories come back together, then their cesium clocks may show different elapsed time. Everything that happened in each laboratory obeyed the same laws of physics in the same way. But, the paths they took through spacetime were of different lengths (different total proper time) - literally, therefore, more proper time passed for one lab than the other.

GR is very much about the nature of time and not the specific workings of clocks.

 

[vanhees71]

The twin paradox is very simple to formulate. You have two time-like world lines with the same initial and final point and compare the proper times
$$\tau_j = \frac{1}{c} \int_{\lambda_{j1}}^{\lambda_{j2}} \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
These two scalar quantities give the proper time passed and thus the aging of each of the twins. Case closed.

 

[vanhees71]

OK thanks. My question was in the same reference frame. Whether cesium 133 is far from any mass or in a very intense gravitational field, the second defined using this cesium in its reference frame will always be exactly the same, is it correct?

Of course within GR the frequency used to define the 2nd has to be measured locally in a local inertial rest frame of the Cs atom to define the second.

 

[martinbn]

I am about 6 feet tall or about 2 m tall. So according to you a m is shorter than a foot?

Not important but 6 feet is about 1.8m.

 

[Kairos]

in a local inertial rest frame of the Cs atom to define the second.

you precision about the inertial rest frame confuses me: the second is supposed to be the same irrespective of wether you are in free fall or static in a gravitational field, provided the caesium clock is present in your reference frame. That's the only important thing, isn't it?

 

[Nugatory]

provided the caesium clock is present in your reference frame.

Those words as written don’t say much, because everything is always in all reference frames. How about “provided the caesium atom is at rest relative to the experimental apparatus and moving inertially”?

 

[jbriggs444]

you precision about the inertial rest frame confuses me: the second is supposed to be the same irrespective of wether you are in free fall or static in a gravitational field, provided the caesium clock is present in your reference frame. That's the only important thing, isn't it?

The English language does not easily describe what is going on.

We are scientists. We deal with measurements. We define time to be what a clock measures. So if a cesium clock measures one second then that was one second of time.

It turns out that there are physical limitations on our ability to compare a second measured by an inertial clock at one velocity against a second measured by an inertial clock at another velocity. One can arrange for the clocks to be co-located at the start time. Or one can arrange for the clocks to be co-located at the stop time. One cannot arrange for both -- for inertial clocks.

So we come up with a coordinate system in which an array of my clocks are synchronized. And we use it to assess the start and end time of your clock. Using this standard of comparison, your clock runs slow compared to my array.

Or you set up an array of your clocks against which you can assess the start and end readings for my clock. Using this standard of comparison, my clock runs slow against your array.

[You may explain this by complaining that my array is not synchronized properly. I can look at your array and levy the same complaint. This is the "relativity of simultaneity"].

It is not that my second is truly longer or shorter than your second. It is that comparing the two of them is not a simple matter.

 

[Kairos]

I understand the first part of your sentence :

“provided the caesium atom is at rest relative to the experimental apparatus

but not the last part

.. and moving inertially”

 

[Kairos]

It turns out that there are physical limitations on our ability to compare a second measured by an inertial clock at one velocity against a second measured by an inertial clock at another velocity

who proposed to determine the second using a moving clock??

 

[jbriggs444]

who proposed to determine the second using a moving clock??

I thought that you were the one who wanted to contemplate two relatively moving clocks.

There is no such thing as "moving" or "at rest". There is only "moving relative to" and "at rest relative to".

 

[Kairos]

There is no such thing as "moving" or "at rest". There is only "moving relative to" and "at rest relative to".

If you see a clock moving, it is moving relative to you... hard to argue

 

[jbriggs444]

If you see a clock moving, it is moving relative to you... hard to argue

So if you want to compare a second measured by that clock against a second measured by your wristwatch, how do you plan to do so?

Personally, I can think of a few ways.

1. I'll grab that clock as it passes by, bring it to a stop and hold it for a few seconds while I compare the tick rates.

2. I'll get a running start and jog alongside the moving clock for a few seconds while I compare the tick rates.

3. I'll get a friend to synchronize clocks with mine and walk slowly to some distance away. I'll note the time when the moving clock passes me and the reading on the moving clock then. He'll note the time when the moving clock passes him and the reading on the moving clock then. We'll compare the difference in our clock readings against the difference in the moving clock readings.

4. I'll get a friend to walk a distance away without first synchronizing our clocks. Then we will exchange light speed signals and synchronize our clocks so that we judge that the speed of light to be the same in both directions.

Special relativity makes a prediction for the results of all four methods.

 

[Kairos]

So if you want to

no I don't want to, and it has nothing to do with the question discussed here

 

[jbriggs444]

no I don't want to, and it has nothing to do with the question discussed here

You are incorrect.

 

[PeroK]

no I don't want to, and it has nothing to do with the question discussed here

Just to clarify things: what is the outstanding question?

 

[Nugatory]

I understand the first part of your sentence :
but not the last part

.. and moving inertially”

Are you asking what "moving inertially" means or why that condition is necessary? It is necessary to exclude perverse setups such as placing the cesium atom and the cycle counting apparatus at different positions in an accelerating spaceship.

 

[Algr]

... is the travel duration shorter because: (1) it contains fewer seconds or (2) the number of seconds is the same for both brothers but the traveler's seconds are shorter?

The brothers will disagree. The brother who traveled will say that the seconds were normal, but there were fewer of them. The stationary brother will perceive the traveler as moving in slow motion, and thus say his seconds were longer.

 

[A.T.]

If you see a clock moving, it is moving relative to you... hard to argue

You can see the same clock as moving or not moving, depending on your own rotation. But that rotation won't affect the clock rates.

 

[jbriggs444]

The brothers will disagree. The brother who traveled will say that the seconds were normal, but there were fewer of them. The stationary brother will perceive the traveler as moving in slow motion, and thus say his seconds were longer.

It is a matter of perspective, of course.

As @Dale and yourself (and I) would view it, taking all of the ticks for the traveling clock and spreading them evenly on the timeline of the stay-at-home clock, there are few remote seconds, each of which is longer. The product comes out to the total stay-at-home elapsed time as it must.

Alternately, one could do the accounting from the traveller's point of view. Spreading the stay-at-home ticks evenly on the traveller's time line, one would find many remote seconds, each of which is extra short. The product comes to the traveller's elapsed time (shorter than the stay-at home elapsed time) as it must.

[You would be correct to be somewhat uncomfortable with this particular accounting from the traveling twin's point of view. The traveling twin is not inertial. There are many choices for non-inertial systems of accounting (coordinate systems or at least "foliations"). There are unexpected behaviors in all of them].

A more standard accounting from the traveller's perspective would start with a very few long ticks of the stay-at-home clock (the outbound trip), a large number of hugely short ticks (the turn-around as the definition of simultaneity sweeps forward across the stay-at-home time line) followed by a very few long ticks of the stay-at-home clock (the return voyage).

 

[Dale]

you precision about the inertial rest frame confuses me: the second is supposed to be the same irrespective of wether you are in free fall or static in a gravitational field, provided the caesium clock is present in your reference frame. That's the only important thing, isn't it?

The BIPM has published a "Mise en Pratique" for each of its units. Here is the one for the second:
https://www.bipm.org/utils/en/pdf/si-mep/SI-App2-second.pdf

It says that "The definition of the second should be understood as the definition of the unit of proper time: it applies in a small spatial domain which shares the motion of the caesium atom used to realize the definition." and further it says "the proper second is obtained after application of the special relativistic correction for the velocity of the atom in the laboratory. It is wrong to correct for the local gravitational field."

I also read that as requiring you to be at rest relative to the atom and physically close to it, but not as requiring either you or the atom to be inertial.

 

[Nugatory]

I also read that as requiring you to be at rest relative to the atom and physically close to it, but not as requiring either you or the atom to be inertial.

I agree with that reading. I also read “small spatial domain” (especially in the context of the previous sentence about negligible non-uniformity of the gravitational field) as implying that the setup can be treated as if gravity is a uniform classical force across the domain; thus there exist frames in which the setup is inertial and at rest. Practically speaking, these are the frames we use for experiments sitting on a lab bench in a lab building except when we are looking specifically for tidal effects.

 

[aperakh]

The problem with your question is the: "are".
There is no "are". That is to say, there is no universal frame of reference by which to decide how long the seconds "really are".
What would you compare a second to as a yardstick to determine how long it is?
The seconds ARE there yardstick. And nothing ever changed sizer relative to itself.
Time, in fact, never actually changes, not in any way that is meaningful to talk about.
It is merely a dimension and what changes is one's perception of the rate at which something is moving through it.

So what what you probably really mean to ask was : "Does the traveling brother perceive there being fewer seconds or does he perceive the seconds as being longer?"

And the answer is: Since there is no meaningful way for him to determine whether his seconds are shorter or longer, he experiences fewer seconds.
That is, his clock will run at such a rate as to count fewer seconds, but they would feel normal to him. While his brother's clock on Earth would seem to run too fast (count shorter seconds).
The brother on earth, watching the moving clock, would also agree that it counted fewer seconds, but to him it would appear as though the moving clock was running slower (ie - counting "longer" seconds).

 

[Nugatory]

The brother on earth, watching the moving clock, would also agree that it counted fewer seconds, but to him it would appear as though the moving clock was running slower (ie - counting "longer" seconds).

That is the conclusion earth-twin might come to (although it would be better to pursue the analogy with distance measurement described earlier in this thread) when they compare clock readings at their reunion.

It is most emphatically not what Earth twin will see if they are watching the moving clock: if the twins are watching each other’s clocks through powerful telescopes or the like, then both will see the other clock running slow on the outbound leg and fast on the inbound leg. When they allow for light travel time between the time the clock ticks and they see the tick, they both calculate the other clock is running slow on both legs.

There is a pretty good explanation in the “Doppler Shift Analysis” section of https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[aperakh]

It is most emphatically not what Earth twin will see if they are watching the moving clock: if the twins are watching each other’s clocks through powerful telescopes or the like, then both will see the other clock running slow on the outbound leg and fast on the inbound leg.

Opps. Misspoke. BOTH twins will perceive the other one's clock as running slower than their own.
But I don't know where you get that there would be a difference beetween the motion being inbound or outbound.
Time dilation is independent of direction. The only thing that matters are the relative speeds, not velocities. Lawrence transforms are scalar, not vectoral.

 

[PeterDonis]

I don't know where you get that there would be a difference beetween the motion being inbound or outbound.

Because the relativistic Doppler effect, which is what determines what each observer actually sees (as opposed to what they calculate after correcting for light travel time), is different for inbound vs. outbound.

Time dilation is independent of direction.

But the Doppler effect is not.

Lawrence transforms

You mean Lorentz transforms.

 

[Herman Trivilino]

That is to say, there is no universal frame of reference by which to decide how long the seconds "really are".

The entire point of SR is that you don't need a universal frame. Any inertial frame will do. As long as the clock is at rest with respect to you it can be used to define the duration of a second.

 

[aperakh]

Because the relativistic Doppler effect, which is what determines what each observer actually sees (as opposed to what they calculate after correcting for light travel time), is different for inbound vs. outbound.

I am afraid you are mistaken. Time dilation is NOT the Doppler effect.
Doppler effect does cause blue and red shifts in light from moving object, but time dilation has nothing to do with it.

You mean Lorentz transforms.

Damn you, auto-correct! lol
Thank you. yes, it is obviously what I meant.

 

[aperakh]

The entire point of SR is that you don't need a universal frame. Any inertial frame will do. As long as the clock is at rest with respect to you it can be used to define the duration of a second.

Yes, he can tell the second for himself (ie - in his frame), which will be different from one observed from another frame.
But there is no universsal frame in which one can say the "real" observation is done. Theya re all equally valid.